Problem: Solve for $x$ : $2x^2 + 6x - 140 = 0$
Explanation: Dividing both sides by $2$ gives: $ x^2 + {3}x {-70} = 0 $ The coefficient on the $x$ term is $3$ and the constant term is $-70$ , so we need to find two numbers that add up to $3$ and multiply to $-70$ The two numbers $-7$ and $10$ satisfy both conditions: $ {-7} + {10} = {3} $ $ {-7} \times {10} = {-70} $ $(x {-7}) (x + {10}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -7) (x + 10) = 0$ $x - 7 = 0$ or $x + 10 = 0$ Thus, $x = 7$ and $x = -10$ are the solutions.